Let the lengths of the bottom of the box be x and y, and let the length of the squares being cu be z, then
V = xyz . . . (1)
2z + x = 16 => x = 16 - 2z . . . (2)
2z + y = 30 => y = 30 - 2z . . . (3)
Putting (2) and (3) into (1) gives:
V = (16 - 2z)(30 - 2z)z = z(480 - 32z - 60z + 4z^2) = z(480 - 92z + 4z^2) = 480z - 92z^2 + 4z^3
For maximum volume, dV/dz = 0
dV/dz = 480 - 184z + 12z^2 = 0
3z^2 - 46z + 120 = 0
z = 3 1/3 inches
Therefore, for maximum volume, a square of length 3 1/3 (3.33) inches should be cut out from each corner of the cardboard.
The maximum volume is 725 25/27 (725.9) cubic inches.
